Problem: $\lim_{x\to\infty}\dfrac{5x-2x^2}{e^{x+3}}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $0$ (Choice C) C $5$ (Choice D) D $-\infty$
Explanation: $\lim_{x\to\infty} 5x-2x^2=-\infty$ and $\lim_{x\to\infty} e^{x+3}=\infty$, so $\lim_{x\to\infty}\dfrac{5x-2x^2}{e^{x+3}}$ results in the indeterminate form $\dfrac{-\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{5x-2x^2}{e^{x+3}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[5x-2x^2\right]}{\dfrac{d}{dx}[e^{x+3}]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{5-4x}{e^{x+3}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[5-4x\right]}{\dfrac{d}{dx}[e^{x+3}]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{-4}{e^{x+3}} \\\\ &=0 \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{-\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[5-4x\right]}{\dfrac{d}{dx}[e^{x+3}]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{5x-2x^2}{e^{x+3}}=0$.